The average of finitely many numbers is their sum divided by how many there are. For a function, which has infinitely many values, the definite integral plays the role of the sum.
Key idea
The average value of $f$ on $[a, b]$ is $$f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx.$$ It is the height of the rectangle on $[a, b]$ with the same area as the region under $f$. If $f$ is continuous, some $c$ in $[a, b]$ has $f(c) = f_{\text{avg}}$ (the Mean Value Theorem for integrals).
Why it works
Sample $f$ at $n$ equally spaced points and average: $\frac{1}{n}\sum f(x_i)$. Since $\Delta x = \frac{b - a}{n}$, we have $\frac{1}{n} = \frac{\Delta x}{b - a}$, so the average is $\frac{1}{b - a}\sum f(x_i)\,\Delta x$, a Riemann sum divided by the length. Letting $n \to \infty$ gives the formula. Because a continuous $f$ takes every value between its min and max, and the average lies between those, $f$ must hit the average somewhere.
Worked example: A polynomial
Find the average value of $f(x) = x^2$ on $[0, 3]$, and where $f$ equals it.
- $\int_0^3 x^2\,dx = \frac{27}{3} = 9$.
- $f_{\text{avg}} = \frac{9}{3} = 3$.
- $c^2 = 3$ gives $c = \sqrt{3}$ in $[0, 3]$.
Answer: $f_{\text{avg}} = 3$, reached at $x = \sqrt{3}$.
Worked example: Sine over an arch
Find the average value of $\sin x$ on $[0, \pi]$.
- $\int_0^{\pi}\sin x\,dx = 2$.
- Divide by the length $\pi$: $\frac{2}{\pi} \approx 0.637$.
Answer: $\frac{2}{\pi}$, a bit less than $\frac{2}{3}$ of the peak height.
Worked example: Average temperature
A model of the temperature (°F) $t$ hours after 6 a.m. is $T(t) = 50 + 4t - \frac{t^2}{4}$. Find the average temperature over the first 12 hours.
- $$\int_0^{12} T\,dt = \Big[50t + 2t^2 - \frac{t^3}{12}\Big]_0^{12} = 600 + 288 - 144 = 744$$.
- Average $= \frac{744}{12} = 62$.
Answer: 62 °F
Common mistake
Averaging the endpoint values, $\frac{f(a) + f(b)}{2}$, instead of integrating. For $x^2$ on $[0, 3]$ that gives 4.5 instead of 3. Endpoint averaging only works for straight lines. Check: the average must lie between the minimum and maximum of $f$ on the interval.